2

这是我的文本文件中的数据：（我已经显示了 10,000 行中的 10 行）索引是行名，temp 是时间序列，m 是以 mm 为单位的值。

     "Index" "temp" "m"
   1 "2012-02-07 18:15:13" "4297"
   2 "2012-02-07 18:30:04" "4296"
   3 "2012-02-07 18:45:10" "4297"
   4 "2012-02-07 19:00:01" "4297"
   5 "2012-02-07 19:15:07" "4298"
   6 "2012-02-07 19:30:13" "4299"
   7 "2012-02-07 19:45:04" "4299"
   8 "2012-02-07 20:00:10" "4299"
   9 "2012-02-07 20:15:01" "4300"
   10 "2012-02-07 20:30:07" "4301"

我使用这个在 r 中导入：

    x2=read.table("data.txt", header=TRUE)

我尝试使用以下代码将时间序列汇总为每日数据：

   c=aggregate(ts(x2[, 2], freq = 96), 1, mean)

我已将频率设置为 96，因为 15 分钟数据 24 小时将包含在 96 个值中。

它返回给我：

    Time Series:
   Start = 1 
   End = 5 
   Frequency = 1 
   [1] 5366.698 5325.115 5311.969 5288.542 5331.115

但我想要与原始数据相同的格式，即我还想要值旁边的时间序列。我需要帮助来实现这一目标。

4

2 回答 2

5

将数据转换为对象后使用包中apply.daily的：xtsxts

像这样的东西应该工作：

x2 = read.table(header=TRUE, text='     "Index" "temp" "m"
1 "2012-02-07 18:15:13" "4297"
2 "2012-02-07 18:30:04" "4296"
3 "2012-02-07 18:45:10" "4297"
4 "2012-02-07 19:00:01" "4297"
5 "2012-02-07 19:15:07" "4298"
6 "2012-02-07 19:30:13" "4299"
7 "2012-02-07 19:45:04" "4299"
8 "2012-02-07 20:00:10" "4299"
9 "2012-02-07 20:15:01" "4300"
10 "2012-02-07 20:30:07" "4301"')

x2$temp = as.POSIXct(strptime(x2$temp, "%Y-%m-%d %H:%M:%S"))
require(xts)
x2 = xts(x = x2$m, order.by = x2$temp)
apply.daily(x2, mean)
##                       [,1]
## 2012-02-07 20:30:07 4298.3

更新：您的问题采用可重现的格式（使用假数据）

我们并不总是需要实际的数据集来帮助解决问题......

set.seed(1) # So you can get the same numbers as I do
x = data.frame(datetime = seq(ISOdatetime(1970, 1, 1, 0, 0, 0), 
                              length = 384, by = 900), 
               m = sample(2000:4000, 384, replace = TRUE))
head(x)
#              datetime    m
# 1 1970-01-01 00:00:00 2531
# 2 1970-01-01 00:15:00 2744
# 3 1970-01-01 00:30:00 3146
# 4 1970-01-01 00:45:00 3817
# 5 1970-01-01 01:00:00 2403
# 6 1970-01-01 01:15:00 3797
require(xts)
x2 = xts(x$m, x$datetime)
head(x2)
#                     [,1]
# 1970-01-01 00:00:00 2531
# 1970-01-01 00:15:00 2744
# 1970-01-01 00:30:00 3146
# 1970-01-01 00:45:00 3817
# 1970-01-01 01:00:00 2403
# 1970-01-01 01:15:00 3797
apply.daily(x2, mean)
#                         [,1]
# 1970-01-01 23:45:00 3031.302
# 1970-01-02 23:45:00 3043.250
# 1970-01-03 23:45:00 2896.771
# 1970-01-04 23:45:00 2996.479

更新 2：替代方法

（使用我在上述更新中提供的虚假数据。）

data.frame(time = x[seq(96, nrow(x), by=96), 1],
           mean = aggregate(ts(x[, 2], freq = 96), 1, mean))
#               time     mean
# 1 1970-01-01 23:45 3031.302
# 2 1970-01-02 23:45 3043.250
# 3 1970-01-03 23:45 2896.771
# 4 1970-01-04 23:45 2996.479
于 2012-06-25T09:58:39.267 回答
2

这将是在基础 R 中执行此操作的一种方法：

x2 <- within(x2, {
   temp <- as.POSIXct(temp, format='%Y-%m-%d %H:%M:%S')
   days <- as.POSIXct(cut(temp, breaks='days'))
   m <- as.numeric(m)
})

with(x2, aggregate(m, by=list(days=days), mean))
于 2012-06-25T11:56:22.120 回答